Mixing
Find the smallest positive integer $ell$ such that $3 cdot left(4^m + 1right)$ divides $2^ell-1$
$begingroup$
Find the smallest positive integer $ell$ such that $3 cdot left(4^m + 1right)$ divides $2^ell-1$
Hint: The sought $ell$ is the multiplicative order of $2$ in the ring of integer residues modulo $3cdot(4^m+1)$.
I am having trouble understanding the hint, are there any "special" characteristics of a ring of integer that might help me?
The answer is supposed to be $4m$ , and I have already managed to show that $3 cdot left(4^m + 1right)$ divides $2^{4m}-1$ , but I do not know how to prove that this is the smallest positive solution.
Link to the question
abstract-algebra number-theory ring-theory coding-theory
edited Jan 15 at 18:55
Mutantoe
612513
asked Jan 15 at 18:36
user635073user635073
84
$endgroup$
add a comment |
$begingroup$
Find the smallest positive integer $ell$ such that $3 cdot left(4^m + 1right)$ divides $2^ell-1$
Hint: The sought $ell$ is the multiplicative order of $2$ in the ring of integer residues modulo $3cdot(4^m+1)$.
I am having trouble understanding the hint, are there any "special" characteristics of a ring of integer that might help me?
The answer is supposed to be $4m$ , and I have already managed to show that $3 cdot left(4^m + 1right)$ divides $2^{4m}-1$ , but I do not know how to prove that this is the smallest positive solution.
Link to the question
abstract-algebra number-theory ring-theory coding-theory
edited Jan 15 at 18:55
Mutantoe
612513
asked Jan 15 at 18:36
user635073user635073
84
$endgroup$
1
$begingroup$
Thanks for the well-asked question! The divisibility you've already established shows that the order must at least divide $4m$. Is it possible for $3(4^m+1)$ to divide $2^ell-1$ if $ellle 2m$?
$endgroup$
– Greg Martin
Jan 15 at 20:15
1
$begingroup$
3(4^m+1)>4^m=2^(2m) , so that mean that 3(4^m+1) can not divide 2^ℓ-1 if ℓ is smaller then 2m , but i still do know how to show that there is no other ℓ between 2m to 4m
$endgroup$
– user635073
Jan 15 at 21:09
$begingroup$
Do you know the following fact? If $a$ divides $2^ell-1$, then the multiplicative order of $2$ in the ring of integer residues modulo $a$ divides $ell$. (The point to emphasize here is that the order must not only be at most $ell$ but must actually divide $ell$.)
$endgroup$
– Greg Martin
Jan 16 at 4:45
add a comment |
$begingroup$
Find the smallest positive integer $ell$ such that $3 cdot left(4^m + 1right)$ divides $2^ell-1$
Hint: The sought $ell$ is the multiplicative order of $2$ in the ring of integer residues modulo $3cdot(4^m+1)$.
I am having trouble understanding the hint, are there any "special" characteristics of a ring of integer that might help me?
The answer is supposed to be $4m$ , and I have already managed to show that $3 cdot left(4^m + 1right)$ divides $2^{4m}-1$ , but I do not know how to prove that this is the smallest positive solution.
Link to the question
abstract-algebra number-theory ring-theory coding-theory
edited Jan 15 at 18:55
Mutantoe
612513
asked Jan 15 at 18:36
user635073user635073
84
$endgroup$
Find the smallest positive integer $ell$ such that $3 cdot left(4^m + 1right)$ divides $2^ell-1$
Hint: The sought $ell$ is the multiplicative order of $2$ in the ring of integer residues modulo $3cdot(4^m+1)$.
I am having trouble understanding the hint, are there any "special" characteristics of a ring of integer that might help me?
The answer is supposed to be $4m$ , and I have already managed to show that $3 cdot left(4^m + 1right)$ divides $2^{4m}-1$ , but I do not know how to prove that this is the smallest positive solution.
Link to the question
abstract-algebra number-theory ring-theory coding-theory
abstract-algebra number-theory ring-theory coding-theory
edited Jan 15 at 18:55
Mutantoe
612513
asked Jan 15 at 18:36
user635073user635073
84
edited Jan 15 at 18:55
Mutantoe
612513
asked Jan 15 at 18:36
user635073user635073
84
edited Jan 15 at 18:55
Mutantoe
612513
edited Jan 15 at 18:55
Mutantoe
612513
edited Jan 15 at 18:55
Mutantoe
612513
612513
asked Jan 15 at 18:36
user635073user635073
84
asked Jan 15 at 18:36
user635073user635073
84
asked Jan 15 at 18:36
user635073user635073
84
84
1
$begingroup$
Thanks for the well-asked question! The divisibility you've already established shows that the order must at least divide $4m$. Is it possible for $3(4^m+1)$ to divide $2^ell-1$ if $ellle 2m$?
$endgroup$
– Greg Martin
Jan 15 at 20:15
1
$begingroup$
3(4^m+1)>4^m=2^(2m) , so that mean that 3(4^m+1) can not divide 2^ℓ-1 if ℓ is smaller then 2m , but i still do know how to show that there is no other ℓ between 2m to 4m
$endgroup$
– user635073
Jan 15 at 21:09
$begingroup$
Do you know the following fact? If $a$ divides $2^ell-1$, then the multiplicative order of $2$ in the ring of integer residues modulo $a$ divides $ell$. (The point to emphasize here is that the order must not only be at most $ell$ but must actually divide $ell$.)
$endgroup$
– Greg Martin
Jan 16 at 4:45
add a comment |
1
$begingroup$
Thanks for the well-asked question! The divisibility you've already established shows that the order must at least divide $4m$. Is it possible for $3(4^m+1)$ to divide $2^ell-1$ if $ellle 2m$?
$endgroup$
– Greg Martin
Jan 15 at 20:15
1
$begingroup$
3(4^m+1)>4^m=2^(2m) , so that mean that 3(4^m+1) can not divide 2^ℓ-1 if ℓ is smaller then 2m , but i still do know how to show that there is no other ℓ between 2m to 4m
$endgroup$
– user635073
Jan 15 at 21:09
$begingroup$
Do you know the following fact? If $a$ divides $2^ell-1$, then the multiplicative order of $2$ in the ring of integer residues modulo $a$ divides $ell$. (The point to emphasize here is that the order must not only be at most $ell$ but must actually divide $ell$.)
$endgroup$
– Greg Martin
Jan 16 at 4:45
1
1
$begingroup$
Thanks for the well-asked question! The divisibility you've already established shows that the order must at least divide $4m$. Is it possible for $3(4^m+1)$ to divide $2^ell-1$ if $ellle 2m$?
$endgroup$
– Greg Martin
Jan 15 at 20:15
$begingroup$
Thanks for the well-asked question! The divisibility you've already established shows that the order must at least divide $4m$. Is it possible for $3(4^m+1)$ to divide $2^ell-1$ if $ellle 2m$?
$endgroup$
– Greg Martin
Jan 15 at 20:15
1
1
$begingroup$
3(4^m+1)>4^m=2^(2m) , so that mean that 3(4^m+1) can not divide 2^ℓ-1 if ℓ is smaller then 2m , but i still do know how to show that there is no other ℓ between 2m to 4m
$endgroup$
– user635073
Jan 15 at 21:09
$begingroup$
3(4^m+1)>4^m=2^(2m) , so that mean that 3(4^m+1) can not divide 2^ℓ-1 if ℓ is smaller then 2m , but i still do know how to show that there is no other ℓ between 2m to 4m
$endgroup$
– user635073
Jan 15 at 21:09
$begingroup$
Do you know the following fact? If $a$ divides $2^ell-1$, then the multiplicative order of $2$ in the ring of integer residues modulo $a$ divides $ell$. (The point to emphasize here is that the order must not only be at most $ell$ but must actually divide $ell$.)
$endgroup$
– Greg Martin
Jan 16 at 4:45
$begingroup$
Do you know the following fact? If $a$ divides $2^ell-1$, then the multiplicative order of $2$ in the ring of integer residues modulo $a$ divides $ell$. (The point to emphasize here is that the order must not only be at most $ell$ but must actually divide $ell$.)
$endgroup$
– Greg Martin
Jan 16 at 4:45
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I may have done this a different way than from suggested in your hint OP.
Claim: $2^{2m}+1 =4^m+1$ does not divide $2^{ell}-1$ for $ell < 4m$.
Proof: $(2^{ell-2m}-1)(2^{2m}+1) = 2^{ell}+2^{ell-2m}-2^{2m}-1 < 2^{ell}-1$ if $ell$ satisfies the inequality $ell -2m < 2m$ or equivalently if $ell$ satisfies the inequality $ell < 4m$. On the other hand $2^{ell-2m} (2^{2m}+1) > 2^{ell}-1$. [Do you see how this implies that $2^{2m}+1$ indeed does not divide $2^{ell}-1$? For some integer $a$,
the integer $a(2^{2m}+1)$ is strictly less than $2^{ell}-1$ yet the integer $(a+1)(2^{2m}+1)$ is strictly greater than $2^{ell}-1$.]
So $ell$ as in your problem must be at least $4m$.
However, $ell=4m$ works; indeed
$$2^{4m}-1 = (2^{2m}-1)(2^{2m}+1) = (2^{2m}-1)(4^m+1).$$
So now it remains to show that $3|(2^{4m}-1)$. However, note that $3|(2^{2m}-1)$ for every integer $m$ [make sure you see why], so $ell=4m$ indeed works; $3times(2^{2m}+1)$ divides $(2^{4m}-1)$.
answered Jan 17 at 23:12
MikeMike
4,151412
$endgroup$
add a comment |
$begingroup$
The ring of integers in the hint can be decomposed by the CRT into the product of $A=Bbb{Z}_3$ by $B=Bbb{Z}_{4^m+1}$. The order of 2 in each of these two rings is respectively 2 (because in $A$, $2^2=1$), and $4m$ (because in $B$, $2^{2m}=-1$) . Hence the order in the product is $operatorname{LCM}(2,4m)=4m$.
edited Jan 17 at 22:51
Jyrki Lahtonen
109k13169374
answered Jan 15 at 23:24
Patrick SolePatrick Sole
1227
$endgroup$
1
$begingroup$
This answer does not prove the assertion that the order modulo $B$ equals $4m$.
$endgroup$
– Greg Martin
Jan 16 at 4:46
1
$begingroup$
If we have in B, the equality x^s=-1that certainly shows the order of x divides 2s and cannot be s, nor a divisor of s. So it has to be 2s.
$endgroup$
– Patrick Sole
Jan 16 at 13:38
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
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$begingroup$
I may have done this a different way than from suggested in your hint OP.
Claim: $2^{2m}+1 =4^m+1$ does not divide $2^{ell}-1$ for $ell < 4m$.
Proof: $(2^{ell-2m}-1)(2^{2m}+1) = 2^{ell}+2^{ell-2m}-2^{2m}-1 < 2^{ell}-1$ if $ell$ satisfies the inequality $ell -2m < 2m$ or equivalently if $ell$ satisfies the inequality $ell < 4m$. On the other hand $2^{ell-2m} (2^{2m}+1) > 2^{ell}-1$. [Do you see how this implies that $2^{2m}+1$ indeed does not divide $2^{ell}-1$? For some integer $a$,
the integer $a(2^{2m}+1)$ is strictly less than $2^{ell}-1$ yet the integer $(a+1)(2^{2m}+1)$ is strictly greater than $2^{ell}-1$.]
So $ell$ as in your problem must be at least $4m$.
However, $ell=4m$ works; indeed
$$2^{4m}-1 = (2^{2m}-1)(2^{2m}+1) = (2^{2m}-1)(4^m+1).$$
So now it remains to show that $3|(2^{4m}-1)$. However, note that $3|(2^{2m}-1)$ for every integer $m$ [make sure you see why], so $ell=4m$ indeed works; $3times(2^{2m}+1)$ divides $(2^{4m}-1)$.
answered Jan 17 at 23:12
MikeMike
4,151412
$endgroup$
add a comment |
$begingroup$
I may have done this a different way than from suggested in your hint OP.
Claim: $2^{2m}+1 =4^m+1$ does not divide $2^{ell}-1$ for $ell < 4m$.
Proof: $(2^{ell-2m}-1)(2^{2m}+1) = 2^{ell}+2^{ell-2m}-2^{2m}-1 < 2^{ell}-1$ if $ell$ satisfies the inequality $ell -2m < 2m$ or equivalently if $ell$ satisfies the inequality $ell < 4m$. On the other hand $2^{ell-2m} (2^{2m}+1) > 2^{ell}-1$. [Do you see how this implies that $2^{2m}+1$ indeed does not divide $2^{ell}-1$? For some integer $a$,
the integer $a(2^{2m}+1)$ is strictly less than $2^{ell}-1$ yet the integer $(a+1)(2^{2m}+1)$ is strictly greater than $2^{ell}-1$.]
So $ell$ as in your problem must be at least $4m$.
However, $ell=4m$ works; indeed
$$2^{4m}-1 = (2^{2m}-1)(2^{2m}+1) = (2^{2m}-1)(4^m+1).$$
So now it remains to show that $3|(2^{4m}-1)$. However, note that $3|(2^{2m}-1)$ for every integer $m$ [make sure you see why], so $ell=4m$ indeed works; $3times(2^{2m}+1)$ divides $(2^{4m}-1)$.
answered Jan 17 at 23:12
MikeMike
4,151412
$endgroup$
add a comment |
$begingroup$
I may have done this a different way than from suggested in your hint OP.
Claim: $2^{2m}+1 =4^m+1$ does not divide $2^{ell}-1$ for $ell < 4m$.
Proof: $(2^{ell-2m}-1)(2^{2m}+1) = 2^{ell}+2^{ell-2m}-2^{2m}-1 < 2^{ell}-1$ if $ell$ satisfies the inequality $ell -2m < 2m$ or equivalently if $ell$ satisfies the inequality $ell < 4m$. On the other hand $2^{ell-2m} (2^{2m}+1) > 2^{ell}-1$. [Do you see how this implies that $2^{2m}+1$ indeed does not divide $2^{ell}-1$? For some integer $a$,
the integer $a(2^{2m}+1)$ is strictly less than $2^{ell}-1$ yet the integer $(a+1)(2^{2m}+1)$ is strictly greater than $2^{ell}-1$.]
So $ell$ as in your problem must be at least $4m$.
However, $ell=4m$ works; indeed
$$2^{4m}-1 = (2^{2m}-1)(2^{2m}+1) = (2^{2m}-1)(4^m+1).$$
So now it remains to show that $3|(2^{4m}-1)$. However, note that $3|(2^{2m}-1)$ for every integer $m$ [make sure you see why], so $ell=4m$ indeed works; $3times(2^{2m}+1)$ divides $(2^{4m}-1)$.
answered Jan 17 at 23:12
MikeMike
4,151412
$endgroup$
I may have done this a different way than from suggested in your hint OP.
Claim: $2^{2m}+1 =4^m+1$ does not divide $2^{ell}-1$ for $ell < 4m$.
Proof: $(2^{ell-2m}-1)(2^{2m}+1) = 2^{ell}+2^{ell-2m}-2^{2m}-1 < 2^{ell}-1$ if $ell$ satisfies the inequality $ell -2m < 2m$ or equivalently if $ell$ satisfies the inequality $ell < 4m$. On the other hand $2^{ell-2m} (2^{2m}+1) > 2^{ell}-1$. [Do you see how this implies that $2^{2m}+1$ indeed does not divide $2^{ell}-1$? For some integer $a$,
the integer $a(2^{2m}+1)$ is strictly less than $2^{ell}-1$ yet the integer $(a+1)(2^{2m}+1)$ is strictly greater than $2^{ell}-1$.]
So $ell$ as in your problem must be at least $4m$.
However, $ell=4m$ works; indeed
$$2^{4m}-1 = (2^{2m}-1)(2^{2m}+1) = (2^{2m}-1)(4^m+1).$$
So now it remains to show that $3|(2^{4m}-1)$. However, note that $3|(2^{2m}-1)$ for every integer $m$ [make sure you see why], so $ell=4m$ indeed works; $3times(2^{2m}+1)$ divides $(2^{4m}-1)$.
answered Jan 17 at 23:12
MikeMike
4,151412
edited Jan 17 at 23:25
answered Jan 17 at 23:12
MikeMike
4,151412
answered Jan 17 at 23:12
MikeMike
4,151412
answered Jan 17 at 23:12
MikeMike
4,151412
4,151412
add a comment |
add a comment |
$begingroup$
The ring of integers in the hint can be decomposed by the CRT into the product of $A=Bbb{Z}_3$ by $B=Bbb{Z}_{4^m+1}$. The order of 2 in each of these two rings is respectively 2 (because in $A$, $2^2=1$), and $4m$ (because in $B$, $2^{2m}=-1$) . Hence the order in the product is $operatorname{LCM}(2,4m)=4m$.
edited Jan 17 at 22:51
Jyrki Lahtonen
109k13169374
answered Jan 15 at 23:24
Patrick SolePatrick Sole
1227
$endgroup$
1
$begingroup$
This answer does not prove the assertion that the order modulo $B$ equals $4m$.
$endgroup$
– Greg Martin
Jan 16 at 4:46
1
$begingroup$
If we have in B, the equality x^s=-1that certainly shows the order of x divides 2s and cannot be s, nor a divisor of s. So it has to be 2s.
$endgroup$
– Patrick Sole
Jan 16 at 13:38
add a comment |
$begingroup$
The ring of integers in the hint can be decomposed by the CRT into the product of $A=Bbb{Z}_3$ by $B=Bbb{Z}_{4^m+1}$. The order of 2 in each of these two rings is respectively 2 (because in $A$, $2^2=1$), and $4m$ (because in $B$, $2^{2m}=-1$) . Hence the order in the product is $operatorname{LCM}(2,4m)=4m$.
edited Jan 17 at 22:51
Jyrki Lahtonen
109k13169374
answered Jan 15 at 23:24
Patrick SolePatrick Sole
1227
$endgroup$
1
$begingroup$
This answer does not prove the assertion that the order modulo $B$ equals $4m$.
$endgroup$
– Greg Martin
Jan 16 at 4:46
1
$begingroup$
If we have in B, the equality x^s=-1that certainly shows the order of x divides 2s and cannot be s, nor a divisor of s. So it has to be 2s.
$endgroup$
– Patrick Sole
Jan 16 at 13:38
add a comment |
$begingroup$
The ring of integers in the hint can be decomposed by the CRT into the product of $A=Bbb{Z}_3$ by $B=Bbb{Z}_{4^m+1}$. The order of 2 in each of these two rings is respectively 2 (because in $A$, $2^2=1$), and $4m$ (because in $B$, $2^{2m}=-1$) . Hence the order in the product is $operatorname{LCM}(2,4m)=4m$.
edited Jan 17 at 22:51
Jyrki Lahtonen
109k13169374
answered Jan 15 at 23:24
Patrick SolePatrick Sole
1227
$endgroup$
The ring of integers in the hint can be decomposed by the CRT into the product of $A=Bbb{Z}_3$ by $B=Bbb{Z}_{4^m+1}$. The order of 2 in each of these two rings is respectively 2 (because in $A$, $2^2=1$), and $4m$ (because in $B$, $2^{2m}=-1$) . Hence the order in the product is $operatorname{LCM}(2,4m)=4m$.
edited Jan 17 at 22:51
Jyrki Lahtonen
109k13169374
answered Jan 15 at 23:24
Patrick SolePatrick Sole
1227
edited Jan 17 at 22:51
Jyrki Lahtonen
109k13169374
edited Jan 17 at 22:51
Jyrki Lahtonen
109k13169374
edited Jan 17 at 22:51
Jyrki Lahtonen
109k13169374
109k13169374
answered Jan 15 at 23:24
Patrick SolePatrick Sole
1227
answered Jan 15 at 23:24
Patrick SolePatrick Sole
1227
answered Jan 15 at 23:24
Patrick SolePatrick Sole
1227
1227
1
$begingroup$
This answer does not prove the assertion that the order modulo $B$ equals $4m$.
$endgroup$
– Greg Martin
Jan 16 at 4:46
1
$begingroup$
If we have in B, the equality x^s=-1that certainly shows the order of x divides 2s and cannot be s, nor a divisor of s. So it has to be 2s.
$endgroup$
– Patrick Sole
Jan 16 at 13:38
add a comment |
1
$begingroup$
This answer does not prove the assertion that the order modulo $B$ equals $4m$.
$endgroup$
– Greg Martin
Jan 16 at 4:46
1
$begingroup$
If we have in B, the equality x^s=-1that certainly shows the order of x divides 2s and cannot be s, nor a divisor of s. So it has to be 2s.
$endgroup$
– Patrick Sole
Jan 16 at 13:38
1
1
$begingroup$
This answer does not prove the assertion that the order modulo $B$ equals $4m$.
$endgroup$
– Greg Martin
Jan 16 at 4:46
$begingroup$
This answer does not prove the assertion that the order modulo $B$ equals $4m$.
$endgroup$
– Greg Martin
Jan 16 at 4:46
1
1
$begingroup$
If we have in B, the equality x^s=-1that certainly shows the order of x divides 2s and cannot be s, nor a divisor of s. So it has to be 2s.
$endgroup$
– Patrick Sole
Jan 16 at 13:38
$begingroup$
If we have in B, the equality x^s=-1that certainly shows the order of x divides 2s and cannot be s, nor a divisor of s. So it has to be 2s.
$endgroup$
– Patrick Sole
Jan 16 at 13:38
add a comment |
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1
$begingroup$
Thanks for the well-asked question! The divisibility you've already established shows that the order must at least divide $4m$. Is it possible for $3(4^m+1)$ to divide $2^ell-1$ if $ellle 2m$?
$endgroup$
– Greg Martin
Jan 15 at 20:15
1
$begingroup$
3(4^m+1)>4^m=2^(2m) , so that mean that 3(4^m+1) can not divide 2^ℓ-1 if ℓ is smaller then 2m , but i still do know how to show that there is no other ℓ between 2m to 4m
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– user635073
Jan 15 at 21:09
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Do you know the following fact? If $a$ divides $2^ell-1$, then the multiplicative order of $2$ in the ring of integer residues modulo $a$ divides $ell$. (The point to emphasize here is that the order must not only be at most $ell$ but must actually divide $ell$.)
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– Greg Martin
Jan 16 at 4:45